Problem: Let $y=\dfrac{\sqrt x}{\cos(x)}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\cos(x)+x\sin(x)}{\sqrt x\cos^2(x)}$ (Choice B) B $\dfrac{\cos(x)+2x\sin(x)}{2\sqrt x\cos^2(x)}$ (Choice C) C $\dfrac{\cos(x)-2x\sin(x)}{2\sqrt x\cos^2(x)}$ (Choice D) D $\dfrac{\cos(x)-x\sin(x)}{\sqrt x\cos^2(x)}$
Explanation: $\dfrac{\sqrt x}{\cos(x)}$ is the quotient of two, more basic, expressions: $\sqrt x$ and $\cos(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left[\dfrac{\sqrt x}{\cos(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}[\sqrt x]\cos(x)-\sqrt x\dfrac{d}{dx}[\cos(x)]}{[\cos(x)]^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\dfrac{1}{2\sqrt x}\cdot\cos(x)-\sqrt x[-\sin(x)]}{\cos^2(x)}&&\gray{\text{Differentiate }\sqrt x\text{ and }\cos(x)} \\\\ &=\dfrac{\cos(x)+2x\sin(x)}{2\sqrt x\cos^2(x)}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\cos(x)+2x\sin(x)}{2\sqrt x\cos^2(x)}$